Thermochemistry and the heat of complete combustion

[DottZakapa]

Homework Statement

Calculate the heat Kcal measured at 25.0°C and 1 atm corresponding to the complete combustion of 4.00 L of Benzene (C6H6 d= 0.780 6/mL) when liquid water forms.

Relevant Equations

thermochemistry

The balanced reaction wil be :
$2C_6H_{6(l)}+15O_{2(g)}->12CO_{2(g)}+6H_2O_(l)$
in order to compute the the standard enthalpy of reaction :
$\Delta H°_{f} H_2O_(l)= -285,8 \frac {KJ}{mol}$;
$\Delta H°_{f} CO_{2(g)}= -393,5 \frac {KJ}{mol}$;
$\Delta H°_{f} C_6H_{6(l)}=49,04 \frac {KJ}{mol}$;

$\Delta H°_{R}= \sum \Delta H°_{f} \text{ products} -\sum \Delta H°_{f} \text{ reagents}= [(-285,8 \frac {KJ}{mol}*6\text{ mol})+(-393,5 \frac {KJ}{mol}*12\text{ mol})]-(49,04\frac {KJ}{mol}*2\text{ mol})=-6531,26 KJ$

converting L to mL:
4,00L=4000 mL
The grams of $2C_6H_{6(l)}$:
4000 mL *0,789 $\frac g {mol}$= 3120 g;

Moles of $2C_6H_{6(l)}$:
$\frac {3120 g}{6*12,01\frac g {mol}+6*1,01\frac g {mol} }$= 39,9 mol;

By proportion the moles of liquid water formed will be :
mol $H_2O_(l)$=$\frac{6 \text{ mol } H_2O_(l)*39,9\text{ mol } 2C_6H_{6(l)}}{2 \text{ mol }2C_6H_{6(l)}}$= 120 mol

The heat , by proportionality :
$\frac{120 \text{ mol } *(-6531,26 KJ)}{6 \text{ mol }}$= -131*10^3 KJ = -3,13*10^4 Kcal

Is it all correct in your opinion?
the book says 3,03 *10 ^4 Kcal
the allowed error is 3% by the professor, where am i wrong?

 

[Chestermiller]

The amount of heat that has to be removed per mole of benzene to hold the temperature constant at 25 C is 6531.26/2 = 3265.63 Kj. For 39.9 moles burned, this comes to 130300 Kj.

 

[DottZakapa]

thanks,ok now i got why is positive, but now still i don't get why the result doesn't match the one on the book 3,03 *10 ^4 Kcal, even yours, divided by 4,184 still not matching

 

[Chestermiller]

thanks,ok now i got why is positive, but now still i don't get why the result doesn't match the one on the book 3,03 *10 ^4 Kcal, even yours, divided by 4,184 still not matching

It's not worth worrying about.

 

[DottZakapa]

well, as i said, at the exam there is an error tolerance of 3%, but this still on the range more or less. Thanks anyway